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the vertex form of $h(x)=x^2 - 14x + 6$ is $h(x)=(x - \boldsymbol{downa…

Question

the vertex form of $h(x)=x^2 - 14x + 6$ is $h(x)=(x - \boldsymbol{downarrow})^2 - \boldsymbol{downarrow}$.

Explanation:

Step1: Complete the square for $x$ terms

Take half of the coefficient of $x$: $\frac{-14}{2} = -7$, square it: $(-7)^2 = 49$. Rewrite the function:
$h(x) = (x^2 - 14x + 49) + 6 - 49$

Step2: Rewrite as vertex form

The first three terms form a perfect square trinomial, simplify the constants:
$h(x) = (x - 7)^2 - 43$

Answer:

$h(x) = (x - 7)^2 - 43$
(The first blank is 7, the second blank is 43)